题目内容
已知sin(α+
)+sinα=-
,则cos(α+
)=
.
π |
3 |
4
| ||
5 |
2π |
3 |
4 |
5 |
4 |
5 |
分析:利用两角和的正弦函数与辅助角公式将已知转化为
sin(α+
)=-
,从而可求得sin(α+
)=-
,再利用诱导公式可求得cos(α+
)的值.
3 |
π |
6 |
4
| ||
5 |
π |
6 |
4 |
5 |
2π |
3 |
解答:解:∵sin(α+
)+sinα
=sinαcos
+cosαsin
+sinα
=
sinα+
cosα+sinα
=
sinα+
cosα
=
(
sinα+
cosα)
=
sin(α+
)
=-
,
∴sin(α+
)=-
.
又α+
=(α+
)+
,
∴cos(α+
)=cos[(α+
)+
]=-sin(α+
)=
.
故答案为:
.
π |
3 |
=sinαcos
π |
3 |
π |
3 |
=
1 |
2 |
| ||
2 |
=
3 |
2 |
| ||
2 |
=
3 |
| ||
2 |
1 |
2 |
=
3 |
π |
6 |
=-
4
| ||
5 |
∴sin(α+
π |
6 |
4 |
5 |
又α+
2π |
3 |
π |
6 |
π |
2 |
∴cos(α+
2π |
3 |
π |
6 |
π |
2 |
π |
6 |
4 |
5 |
故答案为:
4 |
5 |
点评:本题考查两角和的正弦与余弦及辅助角公式,求得sin(α+
)=-
是关键,考查观察分析与运算能力,属于中档题.
π |
6 |
4 |
5 |
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